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Standard XII

Physics

Impulsive Forces in Collision

The velocity ...

Question

The velocity of the wind over the surface of the wing of an aeroplane is 80 ms $^{- 1}$ and under the wing 60 ms $^{- 1}$ . If the area of the wing is 4m $^{2}$ , the dynamic lift experienced by the wing is [ density of air $=$ 1.3 kg. m $^{- 3}$ ]:

3640 N

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7280 N

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14560 N

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72800 N

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Solution

The correct option is B 7280 N
Dynamic lift $= \frac{1}{2} A ρ (v_{2}^{2} - v_{1}^{2})$

$= \frac{1}{2} \times 1.3 \times (2800) \times 4$
$= 7280$ N

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The velocity of the wind over the surface of the wing of an aeroplane is 80 ms−1 and under the wing 60 ms−1. If the area of the wing is 4m2, the dynamic lift experienced by the wing is [ density of air = 1.3 kg. m−3]:

The correct option is B 7280 NDynamic lift =12Aρ(v22−v21) =12×1.3×(2800)×4 =7280 N

The velocity of the wind over the surface of the wing of an aeroplane is 80 ms $^{- 1}$ and under the wing 60 ms $^{- 1}$ . If the area of the wing is 4m $^{2}$ , the dynamic lift experienced by the wing is [ density of air $=$ 1.3 kg. m $^{- 3}$ ]:

The correct option is B 7280 N
Dynamic lift $= \frac{1}{2} A ρ (v_{2}^{2} - v_{1}^{2})$

$= \frac{1}{2} \times 1.3 \times (2800) \times 4$
$= 7280$ N